Question

Mapping genes with recombination frequencies in Drosophila 1. Suppose you know of three linked loci on the Drosophila X chromosome (a, b, and c). Assume that you are starting out with true-breeding stocks that show all three recessive traits and true-breeding stocks that show all three dominant traits. A. If you wanted to do a 3-point test cross to map the three loci, how would you set up the cross? What genotypes do you use for males and females in the testcross? B. How how you derive the genotypes that you are going to use for the testcross? C. What phenotype classes do you expect to see in your testcross progeny? (Don't worry about progeny numbers; just give the progeny classes.) 2. In Drosophila, fused vein (fu), sable body (s), and singled bristle (sn) are recessive alleles of three linked genes (wild type, dominant alleles being fu+, +, and sn+). Note that these genes are listed in alphabetical order, not necessarily according to their relative positions. Trihybrid females were crossed to males showing the three dominant traits, and the progeny phenotypes were: Females: fu+s+ sn+ 1029 Males: fu+s+ sn+ fus+ sn+ fu+ s+ sn fu+ s sn+ fu+ s sn fu s sn+ fus+ sn fussn A. Give the genotypes of the parental females and male. B. Construct a genetic map of the region.

Answer 1

1.
A.

A cross to map three linked loci on the X chromosome would require Trihybrid females and tester males. The trihybrid females will be heterozygous for all three loci, carrying one dominant allele for each of the three loci and one recessive allele. The tester male would be recessive for all three traits.

The genotyes to be used are:

Female: ABC/abc
Male: abc

B.

The Male genotype is present in the pure-breeding line with all three recessive traits. The female parent will be derived by crossing a fly from the true-breeding stocks with all three dominant traits to a fly from the true-breeding stock with all three recessive traits. This fly will be trihybrid.

C.

Phenotype classes expected are:

ABC --> Parental
abc --> Parental
ABc --> Non-Parental
Abc --> Non-Parental
AbC --> Non-Parental
aBc --> Non-Parental
abc --> Non-Parental
abC --> Non-Parental

2.

Cross: Trihybrid Female x Triple dominant Male

Progeny Class	No. of Progeny	Parental/Recombinant	Notes
fu+ s+ sn+	69	Recombinant (Single Crossover)	Crossover between fu and s
fu s+ sn+	321	Parental	-
fu+ s+ sn	17	Recombinant (Double Crossover)	The parental chromosome (fu+ s sn) gained the s+ allele from the second parental chromosome
fu+ s sn+	99	Recombinant (Single Crossover)	Crossover between sn and s
fu+ s sn	307	Parental	-
fu s sn+	21	Recombinant (Double Crossover)	The parental chromosome (fu s+ sn+) gained the s allele from the second parental chromosome
fu s+ sn	91	Recombinant (Single Crossover)	Crossover between sn and s
fu s sn	75	Recombinant (Single Crossover)	Crossover between fu and s
	1000

A.

Base on the progeny class data, the two parental chromosomes in the Trihybrid female are fu s+ sn+ and fu+ s sn.

This is because the progeny classes fu+ s+ sn and fu s sn+ are double crossovers, being the least frequent. These progenies are derived when a double crossover event exchanges the s locus between the parental chromosomes, and the fu s+ sn+ chromosome produces the fu s sn+ gamete while the fu+ s sn chromosome produces the fu+ s+ sn gamete.

Therefore, the genotype of the parental female is: fu s+ sn+ / fu+ s sn
And the genotype of the male parent is: fu+ s+ sn+ / Y

B.

Recombination Frequency between fu and s = (100 * Number of fu-s recombinants)/Total No. of progeny
    = (69+17+21+75)*100/1000 = 182/10 = 18.2%

Recombination Frequency between sn and s = (100 * Number of sn-s recombinants)/Total No. of progeny
    = (17+99+21+91)*100/1000 = 228/10 = 22.8%

Recombination Frequency between fu and sn = (100 * Number of fu-sn recombinants)/Total No. of progeny
    = (69+17+21+75+17+99+21+91)*100/1000 = 41%

The genetic map of the region is:

s sn 18.2m. 22.8 my mo